But why does the 2nd Derivative Test identify max's, min's and saddles points?

Why does the Discriminant let you determine max’s, min’s and saddle points? I was explaining to my students an exam problem on maxes and mins and why they got it wrong. I explained a little bit of WHY the 2nd derivative test works, they said they wished I explained this earlier, because it would have helped them understand and remember the rules for determining max’s, mins and saddle points. So that’s the point of this video: WHY does the discriminant tell you the max’s, min’s and saddle points.

The part about the second derivative being greater or less than zero is perhaps the easiest part of this to understand. Recall back to finding critical points in differential calculus, once you found a critical point, you could tell if it was a max, min or inflection point by looking at it’s second derivative. If the second derivative was positive at the critical point, as it is at this point here, that means the curve is ‘concave up’ at that point. If concave up, then we know we’ll have a minimum at x=a. That’s basically what the first line of the second derivatives test is telling you. If the second derivative with respect to x is greater than 0, then that critical point is a local minimum. The second line states the opposite, if the second derivative is less than 0 , then it’s concave DOWN at this point and f(a,b) is a local maximum. This provides some intuition for lines 1 and 2, so let’s consider this example.

Imagine the surface f(x,y)=x^2+y^2, or easier yet, look at it hear in CP3D. This will have a discriminant of D=4 which is greater than 0, it will be concave up in the x directions at the critical point (0,0) so we know it’s a local min. Throwing in some negative signs, the surface will still have a discriminant of D=4 is greater than 0, and it will be concave down in the x directions at the critical point (0,0) so we know that’s a local max now. But what if we changed it to f(x,y)=x^2-y^2? Now It’s still concave up in the x-direction as shown by the red trace, but it’s concave down in the y direction as shown by the blue trace, and this is no longer a local min, it’s a saddle point. The discriminant for this will be D=2(-2)-[0]^2 =-4-[0]^2 =-4 is less than 0 So the fact that the discriminant is less than 0 tells us it’s a saddle point. And why is the discriminant less than 0? Because the f_xx (a,b) f_yy (a,b) term is negative. And why is f_xx (a,b) f_yy (a,b) negative? Because f_xx (a,b) and f_yy (a,b) have different signs. So that’s the most basic way this discriminant gives you a negative number showing a saddle point. And consider if this f_xx (a,b) f_yy (a,b) term is negative and then you subtract the cross term, which is always positive since it’s squared, you still always end up with a negative D, regardless of the f_xy (a,b) term. But what about that term f_xy (a,b)? Why do we need to consider that? Well, the answer is things get a little more complicated in two dimensions, because now you have cross terms. So let me give you an example here. Consider the surface f(x,y)=x^2+y^2+axy If a=0, this is simply our earlier paraboloid that’s concave up in the x and y directions, and that critical point (0,0) is still a minimum, simple. But look what happens when we start increasing that a term. The critical point (0,0) transforms into a saddle point. So does that mean it’s concave up in the x direction but concave down in the y direction, or vice-versa? Nope, neither. We can see trace out the red line here along the surface in the x direction and you can see it is still concave up. We can see trace out the blue line along the surface in the y direction and you can see it is also still concave up. But even though they are both concave up, you can trace out this purple line along the diagonal which is obviously concave down. So that’s why we need to consider f_xy (a,b). In an equation like this, it basically accounts for how significant that cross term is: If a=0, then D is greater than 0 and (0,0) is a local min based on concavity. If a=2, then D=0 and we don’t have a max, min or saddle. If a=4, then D is less than 0 and (0,0) is a saddle point. That cross term is probably the least intuitive piece but hopefully this gives you a feel for how the discriminant and 2nd derivative test work to identify max’s, min’s and saddle points.